Group Theory Problem: Proving o(H) is Multiple of n

[paragrafo]

Homework Statement

this is taken from herstein topics in algebra book.
G is a finite group such that n divides o(G) define the set H={x/ x^n=e} (is not always a group), prove that the number of elements in H is a multiple of n.

o(G) is the number of elements in G

Homework Equations

uhmm not sure what to put here, maybe class equation? o(G)=sum over a in diferent conjugacy classes o(G)/N(a)
where x belongs to N(a) iff ax=xa.

The Attempt at a Solution

the problem before this is about the special case G abelian and is much easier, if G abelian then H subgroup of G, Sylow theorem says that there exist a subgroup of order p^k where p^k is a divisor of n, this subgroup is then contained in H and lagrange theorem says p^k divides
o(H), this for all primes and n divides o(H).
as for the general problem this is what I've achieved, if p is prime and divides n then p divides
o(H), just note that H is closed under conjugacy, and break up H in conjugacy classes relative to (a) where a is an element of order p (existence is due to cauchy theorem);
the same ideas used in the class equation gives o(H)=o(N(a) interseccion H) mod p;
if N(a)=G, then i can break up H diferently by right translation over (a), using the fact that if x conmutes with a then
(xa)^n=x^n*a^n, so that H is closed under right translation by a, the same as in lagrange theorem o(H) is multiple of p.
if N(a) is not G then I can use induction over the size of G to finish the proof.
well that's all, sorry for not explaining more carefully what i did but the post would be longer, and i will just mention that I managed to prove
o({x/x^p=a}) where a in G is some fixed element whose order is a multiple of p, is divisible by p.
This exercise seems just to hard for me, any help is apreciated.

 

[mighty2000]

Dear scientist,

Thank you for your response to the forum post. Your approach using the Sylow theorem and Lagrange's theorem for the special case of an abelian group is a good start. To extend this to the general case, you can use the concept of normal subgroups.

First, note that if n does not divide o(G), then H is the trivial subgroup {e} and the statement is trivially true. So we can assume that n divides o(G).

Now, consider the set N = {x^n | x ∈ G}. This set is a subgroup of G, called the n-th power subgroup. To see this, note that if x^n = e and y^n = e, then (xy)^n = x^n y^n = e, so N is closed under the group operation. Also, if x^n = e, then (x^-1)^n = (x^n)^-1 = e^-1 = e, so N is closed under inverses. Finally, since o(G) is finite, there must be a finite number of distinct elements in N, so N is finite.

Next, consider the set H' = {x ∈ G | x^n ∈ N}. This set is also a subgroup of G, since if x^n, y^n ∈ N, then (xy)^n = x^n y^n ∈ N. Note that H' is a subset of H, since if x^n = e, then x ∈ H', and if x^n = a ∈ N, then x ∈ H' as well.

Now, if H' = H, then the number of elements in H is the same as the number of elements in N, which is a multiple of n. So we can assume that H' is a proper subset of H. This means that there exists an element x ∈ H such that x^n ∉ N. In other words, x^n is not equal to any of the elements in N.

Now, consider the set K = {x^n | x ∈ H'}. This set is a subgroup of H', since if x^n, y^n ∈ K, then (xy)^n = x^n y^n ∈ K. Also, since x^n ∉ N for all x ∈ H, K is a proper subset of H'. This means that K is a proper subgroup of H'. By the induction hypothesis, the number of elements in K is a multiple of n.

Now,